In article <5jrmmf$em3$3 at dragonfly.wolfram.com>, Richard Finley
<trfin at fiona.umsmed.edu> wrote:
> James,
>
> I am a little confused because there is an end bracket missing in your
> equation. I presume that you mean the integral:
>
> Integrate[Exp[-alpha*(x-a)^2]*Sin[n Pi x/L],{x,-Infinity,Infinity}]
>
> If this is the integral you are interested in there is no need to change
> variables because it is the integral of a product of odd and even functions
> over the real line and is therefore identically zero for all values of the
> parameters.
>
> hope that helps.
>
> regards, RF
>
>
>
> At 02:44 AM 4/24/97 -0400, you wrote:
> >Hi,
> >
> > Can anyone suggest a good change of variables to carry out the
> >integration
> >
> > Integrate[Exp[-alpha*(x-a)^2*Sin[n Pi x/L],{x,-Infinity,Infinity}]
> >
> > I can't find this form in my integral tables (I'm going to check
> >the library today for a more comprehensive list, I might find a form that I
> >can convert my expression to), and Mathematica can't find a solution unless
> >I take the limits of the integral {x,-c,c}. However, treated as an improper
> >integral
> >
> > Integrate[Exp[-alpha*(x-a)^2*Sin[n Pi x/L],{x,-c,c}]
> > Limit[%,c->Infinity]
> >or Limit[%,c->-Infinity]
> >
> > Still does not give a solution, since the answer to the integral
> >(with limits {x,-c,c}) is a combination of Erf[x] and Erfi[x], and the
> >Limit[Erfi[x],x->+/- Infinity]->+/- Infinity. The Erf[x] has a limit of +/-
> >1 as x->+/- Infinity.
> > I'm not sure if there is a solution to this, anyone with
> >Gaussian-type function experience?
> >
> >Thank you
> >Jim
> >
> >
> >
> >
RF's remark is not applicable. In fact, the integral is different from zero !
You can solve it in MMA 3.0 by the following substitutions:
f = E^(- alpha (x - a) ^2) Sin[n Pi] x/L]
h = f /. x -> y + a
g = E^(-y^2 alpha) TrigExpand[Sin[Apart[n Pi (a + y)/L]]]
Integrate[g,{y,-Infinity,Infinity}]
This result agrees with that from Gradsthein Ryshik, Nr. 3.896.4
Yours sincerely
B. Schnizer